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Find all roots of the equation `x^4+2x^3-16x^2-22x+7=0`, if one root is `2+sqrt3` |
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Answer» All coefficients are teal, irrational roots will occur in congugate pairs. Hence another root is `2-sqrt(3)`. `:.` Product of these roots `=(x-2-sqrt(3))(x-2+sqrt(3))` `=(x-2)^(2)-3=x^(2)-4x+1` On dividing `x^(4)+2x^(3)-16x^(2)-22x+7` by `x^(2)-4x+1`, then the other quadratic factor is `x^(2)+6x+y`. Then the given equation reduce in the form `(x^(2)-4x+1)(x^(2)+6x+7)=0` `:.x^(2)+6x+7=0` Then `x=(-6+-sqrt(36-28))/2=-3+-sqrt(2)` Hence the other roots are `2-sqrt(3),-3+-sqrt(2)` |
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